Bourbaki

Proposition 2

Tag 0DY0, permanent. Cite it as [Bourbaki, Tag 0DY0].

Algebra, chapter X, § 6. Utilisation de résolutions non canoniques

The following two diagrams are commutative:

TorA(P,M)→ψ(S,R)H(S⊗AR)TorA(g,f)↓↓H(g~⊗f~)TorA(P′,M′)→ψ(S′,R′)H(S′⊗AR′), \begin{array}{ccc} \mathrm{Tor}^A(P, M) & \xrightarrow{\psi(S, R)} & \mathrm{H}(S \otimes_A R) \\ \mathrm{Tor}^A(g, f) \downarrow & & \downarrow \mathrm{H}(\tilde{g} \otimes \tilde{f}) \\ \mathrm{Tor}^A(P', M') & \xrightarrow{\psi(S', R')} & \mathrm{H}(S' \otimes_A R') , \end{array}
H(HomgrA(R′,E′))→φ(R′,E′)ExtA(M′,N′)H(HomgrA(f~,h~))↓↓ExtA(f,h)H(HomgrA(R,E))→φ(R,E)ExtA(M,N). \begin{array}{ccc} \mathrm{H}(\mathrm{Homgr}_A(R', E')) & \xrightarrow{\varphi(R', E')} & \mathrm{Ext}_A(M', N') \\ \mathrm{H}(\mathrm{Homgr}_A(\tilde{f}, \tilde{h})) \downarrow & & \downarrow \mathrm{Ext}_A(f, h) \\ \mathrm{H}(\mathrm{Homgr}_A(R, E)) & \xrightarrow{\varphi(R, E)} & \mathrm{Ext}_A(M, N) . \end{array}

Let α:L(M)→R,α′:L(M′)→R′,γ:L(P)→S,γ′:L(P′)→S′\alpha : L(M) \to R, \alpha' : L(M') \to R', \gamma : L(P) \to S, \gamma' : L(P') \to S' be morphisms of complexes such that

a∘α=pM,a′∘α′=pM′,b∘γ=pP,b′∘γ′=pP′. a \circ \alpha = p_M, \quad a' \circ \alpha' = p_{M'}, \quad b \circ \gamma = p_P, \quad b' \circ \gamma' = p_{P'} .

By definition, H(g~⊗f~)∘ψ(S,R)\mathrm{H}(\tilde{g} \otimes \tilde{f}) \circ \psi(S, R) is equal to

H(g~⊗f~)∘H(γ⊗α)=H((g~∘γ)⊗(f~∘α)), \mathrm{H}(\tilde{g} \otimes \tilde{f}) \circ \mathrm{H}(\gamma \otimes \alpha) = \mathrm{H}((\tilde{g} \circ \gamma) \otimes (\tilde{f} \circ \alpha)) ,

whereas ψ(S′,R′)∘TorA(g,f)\psi(S', R') \circ \mathrm{Tor}^A(g, f) is equal to

H(γ′⊗α′)∘H(L(g)⊗L(f))=H((γ′∘L(g))⊗(α′∘L(f))). \mathrm{H}(\gamma' \otimes \alpha') \circ \mathrm{H}(L(g) \otimes L(f)) = \mathrm{H}((\gamma' \circ L(g)) \otimes (\alpha' \circ L(f))) .

On the other hand, α′∘L(f)\alpha' \circ L(f) and f~∘α\tilde{f} \circ \alpha are two morphisms from L(M)L(M) into R′R' such that a′∘(α′∘L(f))=pM′∘L(f)=f∘pM=f∘a∘α=a′∘(f~∘α)a' \circ (\alpha' \circ L(f)) = p_{M'} \circ L(f) = f \circ p_M = f \circ a \circ \alpha = a' \circ (\tilde{f} \circ \alpha). By X, p. 49, prop. 3, α′∘L(f)\alpha' \circ L(f) and f~∘α\tilde{f} \circ \alpha are homotopic; analogously γ′∘L(g)\gamma' \circ L(g) and g~∘γ\tilde{g} \circ \gamma are homotopic, and consequently so are (γ′∘L(g))⊗(α′∘L(f))(\gamma' \circ L(g)) \otimes (\alpha' \circ L(f)) and (g~∘γ)⊗(f~∘α)(\tilde{g} \circ \gamma) \otimes (\tilde{f} \circ \alpha) by prop. 3 of X, p. 64. Therefore

H(g~⊗f~)∘ψ(S,R)=H((g~∘γ)⊗(f~∘α))=H((γ′∘L(g))⊗(α′∘L(f))) H(\tilde{g} \otimes \tilde{f}) \circ \psi(S, R) = H((\tilde{g} \circ \gamma) \otimes (\tilde{f} \circ \alpha)) = H((\gamma' \circ L(g)) \otimes (\alpha' \circ L(f)))
=ψ(S′,R′)∘TorA(g,f). = \psi(S', R') \circ \mathrm{Tor}^A(g, f) .

One reasons analogously for the second diagram.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

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Translated with machine help, not yet reviewed by a person.